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OPA569: The sink current/power capability of OPA569

tiper liu
Prodigy 250 points
Part Number: OPA569

Hi, TI Expert:

I have a question about OPA569, about the sink current or sink power capability?

OPA569 can output 2A current, I can undertand it as source current capability, then what's the sink current, the same as output current? how to calculate the sink power?

Thanks

 

  • 0
    TI__Expert 5830 points

    Hello Tiper,

     

    The OPA569 can source and sink capability is +/-2A.  The limitation is in the output swing of the capability of the positive and negative rail versus output current.  This can be seen on the 2 graphs before located on page 5 of the OPA569 datasheet. 

     

    Notice that with an Iout=+/-2A with a supply of 5V the output swing is approximately 140mV away from the positive rail and approximately 200mV away from the negative rail.   The DC load power dissipation of the amplifier is from the output current multiplied by the voltage across the output stage transistor.  The equation for both source or sink are shown below for a DC Load:

     

    DC Load Amplifier Dissipation (Source): Pd(source) = (V+ - Vout)*Iout

    DC Load Amplifier Dissipation (Sink): Pd(sink) = (V- - Vout)*Iout

     

    Bare in mind, that in order to calculate total power dissipation of the amplifier with a DC load, the power dissipation from the quiescent current will also need to be accounted for.  The equation for power dissipation quiescent current is shown below:

     

    PQ=IQ*VS  where VS=VCC-VEE

     

    The total amplifier power dissipation for a DC load is:

     

    Pamp_total = PQ + PDC_Load 

     

    For more information on Power dissipation I have attached a link below to TI precision labs video that covers this topic in more detail.

     

    Best regards,

     

    Errol Leon

    Texas Instruments

    Precision Op Amp Applications

  • 0 in reply to Errol Leon6
    Prodigy 250 points

    Hi, Errol:

    Thanks for your detail reply.

    DC Load Amplifier Dissipation (Sink): Pd(sink) = (V- - Vout)*Iout,

    if I use single power supply, V=-5V and V-= 0V and Load = 2A, Vout= 4V,  the Pd(sink) = 8W ?

    Thanks

    Tiper Liu

  • 0 in reply to tiper liu
    TI__Expert 5830 points
    Hello Tiper,

    The example your example, single power supply, V-=-5V and V+= 0V and Load = -2A, Vout= -4V, meaning that a 2ohm load is present at the output. The Pd(sink) = (-5V - (-4V))*(-2A)= 2W.

    Best regards,

    Errol
  • 0 in reply to Errol Leon6
    Prodigy 250 points

    Hi, Errol:

    Understand, Many thanks to your reply.

    Tiper